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2 years ago

Factorise: 6 x2+ 7x -3

Mathematics

2 answers:

nalin [4]2 years ago

8 0

Factoring a quadratic equation can be defined as the process of breaking the equation into the product of its factors.

$\:❏ \: \: \LARGE{\rm{{{\color{orange}{6 {x }^{2} \: + \: 7x \: - 3}}}}}$

Factorize the equation by breaking down the middle term.

$\large \blue\implies \tt \large \: 6 {x }^{2} \: + \: 7x \: - 3$

Let’s identify two factors such that their sum is 7 and the product is -18.

Sum of two factors = 7 = 9 - 2

Product of these two factors = 9 × (-2) = 18

Now, split the middle term.

$\large \blue\implies \tt \large \:6 {x}^{2} \: + \: 9x \: - \: 2x \: - \: 3$

Take the common terms and simplify.

$\: \\ \large\blue\implies \tt \large \:3x(2x \: + \: 3)\: -1(2x \: + \: 3)$

$\\ \large\blue\implies \tt \large \:(3x \: - \:1 ) \quad \: (2x \: + \: 3) \: = \: 0$

Thus, (3x - 1) and (2x + 3) are the factors of the given quadratic equation.

Solving these two linear factors, we get

$\large\blue\implies \tt \large \:x \: = \: \frac{1}{3} \: \: , \: \: \frac{ - 3}{2} \\$

Alisiya [41]2 years ago

8 0

${6x}^{2} + 7x - 3 \\ \\ {6x}^{2} + 9x - 2x - 3 \\ \\ ( {6x}^{2} + 9x) - (2x + 3) \\ \\ 3x(2x + 3) - 1(2x + 3) \\ \\ (3x - 1)(2x + 3).$

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4 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
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$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

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4 years ago